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3 years ago

Simplify the following expression (3x^2y)^3

Mathematics

1 answer:

kolezko [41]3 years ago

6 0

Answer:

$\large\boxed{\left(3x^2y\right)^3=27x^6y^3}$

Step-by-step explanation:

$\text{Use}\ (ab)^n=a^nb^n\ \text{and}\ (a^n)^m=a^{nm}\\\\\left(3x^2y\right)^3=3^3(x^2)^3y^3=27x^6y^3$

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3 years ago

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3 years ago

During halftime of a basketball game, a sling shot launches T-shirts at the crowd. A T-shirt is launched from a height of 4 f

Sonbull [250]

Answer:

(a) The time the T-shirt takes to maximum height is 2 seconds

(b) The maximum height is 68 ft

(c) The range of the function that models the height of the T-shirt over time given above is $4 + 64\cdot t - 16 \cdot t^{2}$

Step-by-step explanation:

Here, we note that the general equation representing the height of the T-shirt as a function of time is

$h = h_1 + u\cdot t - \frac{1}{2} \cdot g \cdot t^{2}$

Where:

h = Height reached by T-shirt

t = Time of flight

u = Initial velocity = 64 ft/s

g = Acceleration due to gravity (negative because upward against gravity) = 32 ft/s²

h₁ = Initial height of T-shirt = 4 ft

(a) The maximum height can be found from the time to maximum height given as

v = u - gt

Where:

u = Initial velocity = 64 ft/s

v = Final upward velocity at maximum height = 0 m/s

g = 32 ft/s²

Therefore,

0 = 64 - 32·t

32·t = 64 and

t = 64/32 = 2 seconds

(b) Therefore, maximum height is then

$h = 4 + 64\times 2 - \frac{1}{2} \times 32 \times 2^{2}$

∴ h = 68 ft

The T-shirt is then caught 41 ft above the court on its way down

$h = h_1 + u\cdot t - \frac{1}{2} \cdot g \cdot t^{2}$

With u = 64 ft/s

g = 32 ft/s² and

h₁ = 4 ft

The equation becomes

$h =4 + 64\cdot t - \frac{1}{2} \times 32 \cdot t^{2} = 4 + 64\cdot t - 16 \cdot t^{2}$ .

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2 years ago